Question

Write the expression in algebraic form. $$ \sec \left(\sin ^{-1} x\right) $$

Short answer

The short answer is: \(\sec(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}}\).

Step-by-step solution

Define the given expression

We have the expression \(\sec(\sin^{-1}(x))\). Here, \(\sin^{-1}(x)\) denotes the arcsine of x, and \(\sec(\theta)\) is the secant of \(\theta\), which is the reciprocal of the cosine function i.e., \(\sec(\theta) = \frac{1}{\cos(\theta)}\).

Let the arcsine of x be θ

To analyze this expression, we let \(\theta = \sin^{-1}(x)\). By definition, it means that \(\sin(\theta) = x\). Now we have \(\sec(\theta)\) to find.

Use the Pythagorean identity

Since we know \(\sin(\theta) = x\), we can use the Pythagorean identity: \(\sin^2(\theta) + \cos^2(\theta) = 1\). We want to find \(\cos(\theta)\), so we will isolate it in the Pythagorean identity: \(\cos^2(\theta) = 1 - \sin^2(\theta)\) Now we substitute \(\sin(\theta) = x\): \(\cos^2(\theta) = 1 - x^2\) Take the square root of both sides to get: \(\cos(\theta) =\pm \sqrt{1-x^2}\) Since θ is the arcsine function of x, and the arcsine function has a range of \([- \frac{\pi}{2}, + \frac{\pi}{2}]\), the cosine function will always be positive within this range. This means that we can ignore the negative square root and just use the positive value: \(\cos(\theta) = \sqrt{1-x^2}\)

Finding the secant

Recall that the secant is the reciprocal of the cosine function. Thus, to find \(\sec(\theta)\), we will take the reciprocal of \(\cos(\theta) = \sqrt{1-x^2}\). So we have: \(\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{\sqrt{1-x^2}}\) With \(\theta = \sin^{-1}(x)\), the algebraic expression becomes: \(\sec(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}}\)